Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)

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- Oct 13th 2009, 10:05 AMCoda202Isomorphic
Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)

- Oct 13th 2009, 10:47 AMeeyore
I'll give it a shot. Assume $\displaystyle \phi: \mathbb{R} \rightarrow \mathbb{Q}$ is such an isomorphism. Say $\displaystyle \phi(1)=k$ where $\displaystyle k \in \mathbb{Q}$. $\displaystyle \phi$ must satisfy

$\displaystyle \phi(ab) = \phi(a)\phi(b)$

so

$\displaystyle kab = k^2 ab \Rightarrow k = 1$

Thus, each integer, and in turn each rational, is mapped according to the identity mapping. Then where would the irrationals go? The homomorphism seems to be onto, but not one-to-one, so no bijection can exist. - Oct 13th 2009, 10:51 AMTaluivren
Hi,

here's my try:

sentence

$\displaystyle \forall x \exists y \; x=y\cdot y\cdot y$

holds in $\displaystyle \mathbb{R}$ but not in $\displaystyle \mathbb{Q}$. So they are not elementarily equivalent, therefore they can't be isomorphic. - Oct 13th 2009, 12:08 PMclic-clac
Another one: assume $\displaystyle f:\mathbb{R}\rightarrow\mathbb{Q}$ is a ring isomorphism, then $\displaystyle f(\sqrt{2})^2=f(\sqrt{2}^2)=f(2)=f(1+1)=f(1)+f(1)= 1+1=2$.

But no rational is a squared root of 2, so $\displaystyle f$ cannot exist.

(assume $\displaystyle p,q$ belong to $\displaystyle \mathbb{N}^*, gcd(p,q)=1$ and $\displaystyle (\frac{p}{q})^2=2$ then $\displaystyle p^2=2q^2$.

But this is impossible since $\displaystyle \nu_2(p^2)$ is even and $\displaystyle \nu_2(2q^2)$ is odd (where $\displaystyle \nu_2(n)$ is the greatest $\displaystyle k$ such that $\displaystyle 2^k$ divides $\displaystyle n$))